Question

A slightly conical wire of length L and end radii $r_1$ and $r_2$ is stretched by two forces applied parallel length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the elongation in the wire is

• A. $\frac{Fl}{2\pi r_1{r}_{2}Y}$
• B. $\frac{Fl}{\pi r_1{r}_{2}Y}$
• C. $\frac{2Fl}{2\pi r_1{r}_{2}Y}$
• D. None.

Answer 1

The correct option is A $\frac{Fl}{2\pi r_1{r}_{2}Y}$

Consider small elements dIsk of radium $r$, of thickness and located at distance $x$ Let elongation of thin small elements be $\delta x$

From the figure we can white,

$\tan \theta =\frac{r-r_1}{x}=\frac{r_2-r_1}{L}$

$\Rightarrow r=\frac{r_2-r_1}{L}x+r_1$ (Radius of elements)

From Hooke's law, elongation of differential elements,

$\delta =\frac{FL}{AY}$ $Y$:Young's modules

$A$:Gross-sectional area.

$\Rightarrow \delta x=\frac{Fdx}{Y\pi \left[\right(\frac{r_2-r_1}{L})x+r_1{]}^2}$

For net elongation, we intergrate $x$ from $o\Rightarrow L$

$\Rightarrow △x=\int \delta x={\int }_0^L\frac{Fdx}{Y\pi \left[\right(\frac{r_2-r_1}{L})x+r_1{]}^2}$

which directry gives:

$△x=\frac{FL}{2\pi r_1{r}_{2}Y}$